Talk:Busy beaver function/Archive 1
Is \(f_{\omega^\text{CK}_1}(n)\) a well-defined function? How to evaluate \(f_{\omega^\text{CK}_1}(2)\), for example? Ikosarakt1 (talk) 11:52, January 28, 2013 (UTC) :We can't, it's uncomputable xD :I have no idea what the fundamental sequence for \(\omega^\text{CK}_1\) is — probably different functions of \(\omega\) stepping up through different levels of recursion. That is, if a fundamental sequence exists. FB100Z • talk • 20:40, January 30, 2013 (UTC) ::Or maybe \(f_{\omega^\text{CK}_1}\) will have to be defined backwards from \(\Sigma\). I dunno. FB100Z • talk • 21:39, February 1, 2013 (UTC) I found site which conjectures definition of \(f_{\omega^\text{CK}_1}(n)\). If it is really that, \(f_{\omega^\text{CK}_1}(2) = f_{\omega}(2) = 2^2 \times 2 = 8\). Ikosarakt1 (talk) 09:05, February 2, 2013 (UTC) \(\Sigma(5)\) Probably, we soon find out whether \(\Sigma(5) = 4098\) or not. According to this list, only 35 machines remains undecided. Ikosarakt1 (talk) 10:07, February 22, 2013 (UTC) Unfortunately, that list is over a decade old and no one has really made progress since. This doesn't seem to be the sort of problem that interests many mathematicians. (There is one J Harland who mentioned he was working towards solving this, but again that was back in 2009.)Deedlit11 (talk) 15:00, February 28, 2013 (UTC) Mathematicians don't care about us *sob* FB100Z • talk • 07:10, March 5, 2013 (UTC) Grounding Nice explanation, FB100Z. I want more 02:51, March 5, 2013 (UTC) That was my work, read history. Ikosarakt1 (talk) 06:29, March 5, 2013 (UTC) :Aw, you ruined my moment of stolen glory :P FB100Z • talk • 07:04, March 5, 2013 (UTC) :Oops! I didn't notice that. Anyway, thanks for improving the explanation, FB100Z! — I want more 08:22, March 5, 2013 (UTC) I'm going to add oracle Turing machines to the informal explanation, but I haven't found a definition that makes sense and is reasonably concrete. Anyone wanna help? FB100Z • talk • 18:47, March 5, 2013 (UTC) I believe that oracle Turing machine looks like that: we have two tapes, two rulesets and oracle box: normal and oracle. Then oracle box can take arbitrary Turing machine program (which is prescribed in the oracle ruleset), and test whether it halts or not. If not, then write nothing at the current oracle cell blank or write 1 otherwise, move into the direction according to the oracle ruleset. Ikosarakt1 (talk) 21:06, March 5, 2013 (UTC) Nice construction, Ikosarakt. I had conceived of an oracle as something like the following: it's a Turing machine with one tape and one ruleset, however, there is a special oracle state (we'll call it state O). When in the oracle state, the Turing machine does not check whether the currect cell is 1 or 0, but rather counts the number of consecutive 1's starting from the currect cell and moving to the left. If that number is n, the oracle will instantly check whether the nth Turing machine halts or doesn't halt. If it halts, one set of instructions (go to state, write either 0 or 1, move either right or left) is executed; if it doesn't halt, another set of instructions is executed. I think this is powerful enough to define an oracle machine, although I haven't proven it. Deedlit11 (talk) 20:44, March 6, 2013 (UTC) Number of consecutive 1's can be very large and larger than the number of TM's with some number of states. For example, there exists 20736 TM's with 2 states, but if there will be, say, 30000 consecutive 1's, we need to check non-existed TM, thus the function freeze. Ikosarakt1 (talk) 22:24, March 6, 2013 (UTC) The oracle in my oracle Turing machine is not limited to 2-state or n-state Turing machines; it can test a Turing machine with any number of states (limited of course by the maximum number of 1's the oracle Turing machine can print out). It's actually important that the oracle can test Turing machines with arbitrarily many states. If your n-state oracle Turing machine can only test Turing machines with n states or less, than it can't evaluate BB(n+1), for example. So it can't evaluate the general function BB(m), much less BB^m(m) and higher functions. So your oracle Turing machine is far less powerful than it needs to be. Remember, we need to be able to evaluate any algorithm where functions Halt(m) and BB(m) are allowed. Deedlit11 (talk) 23:17, March 6, 2013 (UTC) Multi-head TMs Let reconstruct TM such that head moves instead of tape. Then we can add more heads and for each head specify a separate ruleset. Initially all heads occupy one position. Define \(\Sigma(a,b,c)\) to be largest finite number of non-zeros that can be written on the a-state, b-color, c-head Turing machine. Ikosarakt1 (talk ^ 10:52, March 15, 2013 (UTC) :Since a busy beaver uses up a finite amount of space, we can place the machines far apart from each other so they don't interact during their computations. This means that in general \(\Sigma(a, b, c \cdot d) \geq \Sigma(a, b, c) \cdot d\). Also \(S(a, b, c) \geq S(a, b)\). FB100Z • talk • 18:24, March 18, 2013 (UTC)